I came up with my own solution that made perfect sense to me, but when I read the text's solution, it argued that for the primes that are of the particular form are $6k+1 = 3(2k)+1$. But doesn't that really say the primes in the form of $3k+1$ are in the form of $6m+1$? It seems to me as though there's some misuse of notation here -- allowing $k = 2k$. So should the exercise be phrased as $6m+1$ instead?

4 Answers
4

to distinguish the integers $n, k$, and allow for subsequently proving this is the case for $n = 2k$: primes of the form $3n + 1 = 3(2k) + 1$ are thus of the form $6k + 1$.

But just like indexing variables, I suspect that "$k$" as used in the actual problem statement was intended to be a "dummy" variable standing in for "some integer", much like $x$ in the expressions "$\forall x P(x)$, and $\forall x Q(x)$" each use $x$ independently of its use in the corresponding assertion. But this is not standard.

Yeah, but when they used $k$ in both, it reads that the $k$'s are going to be the same. Is this sort of thing standard notation?
–
AlanHMay 3 '13 at 22:55

1

The more standard notation would be that of congruence modulo an integer and so 6n+1=3(2k)+1 would more readily be read as $x=1 \mod 6\Rightarrow x=1 \mod 3$
–
Daniel RustMay 3 '13 at 23:00

I agree with you, AlanH. That's why I suggested the highlighted statement and find it preferable. The problem statement as you posted is ambiguous, perhaps even misleading. I agree with @Daniel Rust, wholeheartedly.
–
amWhyMay 3 '13 at 23:01

The only $k$ for which $3k+1=6k+1$ is $k=0$, which doesn't even give a prime. The statement means "if a prime is a multiple of three plus one, then it is a multiple of six plus one". The $k$ is a dummy variable; although it is not proper to us $k$ twice, the statement is not much clear if you use two different variables.

$3n+1=6k+1$ iff $n=2k.$ If $n$ is even, this will always work, but if $n$ is odd, then $3n+1$ is even and thus not prime and there are no solutions and so the statement is true.
You can of course extend this to the rational numbers, and then it gives every prime, but that's not very interesting.